Let S be the set of all real numbers and Let R be a relations on s defined by `A R B hArr |a|le b.` then ,R is

Options

A. (a)Reflexive
B. (b)Symmetric
C. (c)Transitive
D. (d)Equivalence

Explanation

To determine the properties of the relation \( R \) defined on the set \( S \) of all real numbers by \( A R B \) if and only if \( |A| \leq B \), we will check whether the relation is reflexive, symmetric, and transitive.

### Step 1: Check for Reflexivity
A relation \( R \) is reflexive if for every element \( A \in S \), \( A R A \) holds true. This means we need to check if \( |A| \leq A \).

- Let \( A = -3 \):
\[ |A| = |-3| = 3 \]
We need to check if \( 3 \leq -3 \), which is false.

Since we found a counterexample, the relation is **not reflexive**.

### Step 2: Check for Symmetry
A relation \( R \) is symmetric if whenever \( A R B \) holds, then \( B R A \) must also hold. This means we need to check if \( |A| \leq B \) implies \( |B| \leq A \).

- Let \( A = 4 \) and \( B = 6 \):
\[ |A| = |4| = 4 \quad \text{and} \quad B = 6 \]
We check if \( 4 \leq 6 \) (true), but now we check \( |B| \leq A \):
\[ |B| = |6| = 6 \quad \text{and} \quad A = 4 \]
We check if \( 6 \leq 4 \) (false).

Since we found a counterexample, the relation is **not symmetric**.

### Step 3: Check for Transitivity
A relation \( R \) is transitive if whenever \( A R B \) and \( B R C \) hold, then \( A R C \) must also hold. This means we need to check if \( |A| \leq B \) and \( |B| \leq C \) implies \( |A| \leq C \).

Assume \( |A| \leq B \) and \( |B| \leq C \). We want to show that \( |A| \leq C \).

From \( |A| \leq B \) and \( |B| \leq C \):
- Since \( |A| \leq B \) and \( B \leq C \), by the transitive property of inequalities, we have:
\[ |A| \leq C \]

Thus, the relation is **transitive**.

### Conclusion
The relation \( R \) is:
- Not reflexive
- Not symmetric
- Transitive

Since it is not reflexive or symmetric but is transitive, the relation \( R \) is classified as **transitive only**.